Archimedes' Principle - AP Physics 2

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Question

How much of an iceberg is submerged below the water if the density of ice is $917 \frac{\textup{kg}}{\textup{m}^3}$ and the density of water is $1000 \frac{\textup{kg}}{\textup{m}^3}$ ?

Answer

The volume of the submerged mass is the volume of the mass proportional to the density of the solid to the density of the fluid.

$V_{sub}= V_{solid}(\frac{\rho_{solid}}{\rho_{fluid}})$

Determine the amount of ice that is submerged.

$\frac{V_{sub}}{V_{solid}}= \frac{\rho_{solid}}{\rho_{fluid}}= \frac{917}{1000} = 91.7 %$

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Question

Which of the following is a statement of Archimedes' principle?

Answer

Archimedes' Principle states: When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of fluid displaced by the body. Thus, the buoyant force is dependent on the density of the liquid and volume of the immersed object, regardless of the object's mass or surface area.

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Question

A spherical ball of density $\rho =0.70 \frac{kg}{L}$ has a radius of . If the ball is placed on the surface of water and released, how much of the ball becomes submerged in the water?

$g = 10\frac{m}{s^2}$

Answer

We can use Archimedes's Principle to solve this problem which states that the upward buoyant force on an object is equal to the weight of the fluid that the object displaces. Therefore, if an object is floating, the upward buoyant force is equal to the weight of the object. So, let's begin by calculating that. We are given the radius of the object and told that it is a sphere. Therefore, we can use the expression for the volume of a sphere:

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi (10cm)^3 = \frac{4000\pi}{3}cm^3$

$V = \frac{4000\pi}{3}cm^3\left ( \frac{1L}{1000cm^3}\right ) = \frac{4\pi}{3}L$

Now multiplying this by the density, we get:

$m = \left ( \frac{4\pi}{3}L\right )\left ( 0.70\frac{kg}{L}\right ) = 2.93kg$

This is also going to be the mass of the water displaced. We can therefore calculate the volume of water displaced:

$V = \frac{m}{\rho} = \frac{0.9\bar{3}kg}{\left ( 1.0\frac{kg}{L}\right )} = 2.93L$

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Question

If a object with a volume of $1:m^{3}$ is submerged below water, what is the net force acting on the object?

$\rho_{water}=1000: \frac{kg}{m^{3}}$ .

Answer

To answer this question, we'll need to consider the object submerged under water, and then use a force diagram in order to see which forces are acting on it.

The forces that are relevant to this question are the vertical forces, or those acting in the y-direction. Pointing down, we have the force due to gravity, which is the object's weight. Pointing upward is the bouyant force.

$\Sigma F_{y}=F_{B}-W$

Next, we can expand each of the variables in the above expression.

$\Sigma F_{y}=\rho Vg-mg$

$\Sigma F_{y}=(1000: \frac{kg}{:m^{3}})(1: m^{3})(9.8: \frac{m}{s^{2}})-(100: kg)(9.8: \frac{m}{s^{2}})$

$\Sigma F_{y}=8820: N$

Note that the depth in which the object was submerged is extraneous information.

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Question

A ball of mass is lightly dropped into a tub with a base of . After it sinks to the bottom, the water rises by . Determine the density of the ball.

Answer

The volume displaced will equal the volume of the ball.

$\rho=\frac{m}{V}$

$\rho=\frac{4 kg}{.0025m^3}$

$\rho=\frac{1600kg}{m^3}$

$\frac{1600kg}{m^3}\frac{1m^3}{100^3cm^2}\frac{1000cm^3}{1L}=1.6\frac{kg}{L}$

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